BIO207H5 Lecture Notes - Lecture 15: Centimorgan, Gamete, Meiosis

46 views3 pages

grayprairie-dog650

23 Mar 2017

School

Department

Course

Professor

For unlimited access to Class Notes, a Class+ subscription is required.

Document Summary

The relationship between recombination frequency and map distance can be used to predict frequencies of recombinant gametes based on map distances. If an organism ab/ab has genes a and b 10 map units (or 10%) apart, two parental gametes and two nonparental gametes will be produced: recombinant gametes are expected 10% of the time. Gamete frequencies from genetic maps: both recombinant gametes (a b and a b) should be observed equally frequently, (0. 10) . The parental gametes will be observed when there is no crossover between the genes, or. 100% 10% 90% of the time. The two gametes, a b and a b, will be seen in equal proportion, in this case (0. 90) 0. 45, or 45% each. Experimental evidence shows that map distances calculated between two genes underestimate the physical distance between them: not all crossovers between genes are detected, e. g. , a two-strand double crossover yields only parental allele combinations.

Related Questions

Short answers please

1. What is the ultimate effect in an XY human with its SRY genemutated?

2. In humans, the allele for brown eye (B) is dominant to blue(b) eye color. If two heterozygous parents produce children, whatis the probability that:

a. The first child has brown eyes, and the next two childrenhave blue eyes?

b. If the couple above eventually produces a total of sixchildren, what is the probability that three of them will be boyswith brown eyes, and three of them girls with blue eyes?

3. In a paternity dispute, a type AB woman claimed that one ofthe four men was the father of her type A child. Which of thefollowing men could be the father of the child on the basis of theevidence below? Give your reasoning.

Man #1 = type A

Man #2 = type B

Man #3 = type O

Man #4 = type AB

4. Several mating were made among all-white cats. The progentfrom these mating showed the following phenotypic frequency:

All-white cats = 12/16

Black cats = 3/16

Gray cats = 1/16

a. Diagram the cross showing the genotypes of the parents, andtheir progeny as given above,

b. Which gene is epistatic? Explain your answer.

5. Mitosis and meiosis differ in that

a. homologues are seperated in prophase of mitosis

b. homologues are seperated in prophase I of meiosis

c. homologues are seperated in prophase II of meiosis

d. sister chromatids are seperated in prophase I of meiosis

6. The phases of meiosis that most resemble those of mitosisare

a. the meiosis I phase

b. the meiosis II phase

c. meiosis I, except for the pairing of homologues

d. meiosis II, except for the pairing of homologues

7. Genetic variability in sexually reproducing organism s arisesfrom

a. crossing over events between sister chromatids

b. crossing over events between nonsister chromatids

c. egg and sperm genetic recombination from distinctindividuals

d. egg and sperm recombination from an individual

8. A female contains an extra X chromosome, resulting fromnondisjunction. When this gamete is fertilized with normal sperm,it would result in an individual of a genotype

a. XXX

b. XXY

c. XYY

d. all of the above

e. only a and b

9. A diploid organism has the genotype AaBbCcDdEe. How manydifferent types of gametes can this individual produce?

a. 5

b. 10

c. 25

d. 32

e. 625

10. Assuming complete dominance, what is the phenotype ratio ofthe offspring of the cross AaBb x aabb?

a. 1:1

b. 3:1

c. 1:2:1

d. 1:1:1:1

e. 9:3:3:1

byzantiumkangaroo348

ANSWER ALL the PARTS you can

PART1 Alleles are considered linked if recombination is less than 50% in:

a) a Px P Fl cross

b) an F1 x F1F2 cross

c) an F2 x F2F3 cross

d) an Fl x homozygous recessive cross

e) any F2 progeny x homozygous recessive cross 1.

PART2 If two loci A/a B/b are 9 map units apart, and an Ab/aB individual is testcrossed,

a) There will be 9% AB individuals

b)There will be 91% Ab individuals

c) There will be 4.5% ab individuals

d) There will be 91% recombination individuals

If two genes a and b show 50% recombination genes a and c show 35% recombination, and genes b and show 32% recombination, then

a) Genes a and b are on the same chromosome, and 50 m.u. apart

b) Genes a and b are on different chromosomes

c)Genes a and b are on the same chromosome, and 67 map units apart

d) One cannot tell if genes a and b are on the same or different chromosomes from given data.

PART4 If two genes a and b are 8.3 m.u. a part, one expects to find

a) 8.3% recombinant gametes from a double heterozygous parent

b) 91.7% recombinant gametes from a double heterozygous parent

c) A chiasma between the a and b loci in 8.3% of the meiosis

d) A chiasma between the a and b loci in 91.7% of the meiosis.

PART5 Map distances are most accurate when genes are closely linked because as genes become further apart,

a) They always show independent assortment

b)The chance of multiple crossovers increases

c) They no longer recombine

d) There is chiasma interference

PART6 In a three-point testcross, the most frequent classes of progeny were ABc and abC. The less frequent classes of progeny were AbC and aBc. Which gene is in the middle?

a) Aa

b) B/b

c)C/c

d) It is not possible to tell from just these data

PART7 In a three-point testcross with 100 progeny, there were 16 single crossovers between genes A and B, 29 single crossovers between genes B and C, and 4 double crossovers. The correct gene order was determined to be A-B-C. What is the map distance between A and B?

a) 20%

b) 0.2 m.u

c) 20 m.u.

d) 16 m.u.

e) 29 m.u.

f) 16%

Which of these statements is incorrect?

Syntenic genes are located on the same chromosome.

Independent assortment results in recombinant chromosomes.

You can reliably predict the relative genetic distance fromgenesâ physical distance on a chromosome.

Linked genes are always syntenic.

What is the relative genetic distance between two linked genesif the recombination frequency is 0.49?

0.49 cM

4.9 cM

49 cM

490 cM

What statement best explains the distortion in Mendelian ratiosobserved by Bateson & Punnett in 1905? (Reminder: they found anoverrepresentation of F2 offspring showing both dominant orrecessive phenotypes, and an underrepresentation of offspringdisplaying one dominant and one recessive phenotype)

Human error: they should have been more careful about theirexperimental setup.

Gene linkage: Genes for flower color and pollen shape arephysically close on the same chromosome, leading to a breakdown inthe independent assortment of the alleles for these traits.

Chromosome crossover: Homologous recombination of twochromatids during meiosis caused the alleles to shuffle, resultingin a breakdown of the independent assortment of the alleles forthose genes.

Random variation: No two situations are alike. In finitepopulations, you are going to get some variation across a mean.

When determining the relative genetic distance between twogenes, why is dihybrid back-cross preferable over traditionaldihybrid cross?

9:3:3:1 phenotypic ratio is easier to work with than 1:1:1:1ratio.

Genotypes of the offspring can be determined based on theirphenotype.

If the genes are independently assorted, the dihybrid back-crosswould result in only 2 genotypes in the F1 generation.

B and C

Why do we map genes?

To understand how genes interact with each other

Comparative genomics analysis